Question

5- A 2.0 kg block of aluminum (specific heat = 897 J/kg·K) is at an initial temperature of 300 K. What will its final temperature be if (3.35 × 105 J )of thermal energy is transferred to the block?

Answer 1

Answer:

T = 486.6 K

Explanation:

The final temperature of the block can be found using the following formula:

$Q = mC\Delta T$

where,

Q = Thermal Energy Transferred = 3.35 x 10⁵ J

m = mass of aluminum block = 2 kg

C = Specific Heat = 897 J/kg.K

ΔT = Change in Temperature = T - 300 K

T = Final Temperature of the Block = ?

Therefore,

$3.35\ x\ 10^5\ J = (2\ kg)(897\ J/kg.K)(T-300\ K)\\\\5.38\ x\ 10^5\ J + 3.35\ x\ 10^5\ J = (1794 J/K)(T)\\\\T = \frac{8.73\ x\ 10^5\ J}{1794\ J/K}$

T = 486.6 K